 Solution

Let $P=\{w\in L^2(\Omega ):w\ge 0\text{ almost everywhere}\}$. For $g\in L^2(\Omega )$, define $Gg=z\in H_0^1(\Omega )$ byThe Lax-Milgram theorem guarantees a unique $z$: the bilinear form on the left is bounded and coercive in the norm $\Vert \nabla \cdot {\Vert }_2$, and the right side is a bounded linear functional by the Poincaré inequality. Testing with $z$ givesDefine $T:P\to L^2(\Omega )$ by $Tw=G(1+\sqrt{w})$. This actually maps into $P$. Indeed the weak maximum principle says that a weak solution $z\in H_0^1(\Omega )$ of $-\Delta z=g\ge 0$ is nonnegative almost everywhere. Here $1+\sqrt{w}\ge 0$ and belongs to $L^2(\Omega )$. Directly, taking the negative part $z^{-}=\max \{-z,0\}$ as a test function gives $-\Vert \nabla z^{-}{\Vert }_2^2={\int }_{\Omega }g{z}^{-}dx\ge 0$, so $z^{-}=0$ by the Poincaré inequality.

The required fixed point formulation isEvery such fixed point belongs automatically to $H_0^1(\Omega )$ and satisfies part (b)(i). Conversely, uniqueness in the Lax-Milgram theorem makes every nonnegative weak solution a fixed point of $T$.